Kinetics (Advanced)

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For on-line tutorial help with kinetics, there are four modules available in the Kinetics section of chemcal and in particular, the module Kinetics and Reaction Mechanisms.


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Q: 1 2 3 4 5 6 7


For the reaction represented by 2A + B → C it is found that doubling the concentrations of both A and B quadruples the rate of reaction at a fixed temperature. Doubling the concentration of B alone does not affect the rate of reaction.

(a) Write an expression for the rate law of the reaction

(b) Suggest a sequence of steps compatible with this behaviour.


A probable mechanism for the reaction -

NO2(g) + CO(g) → CO2(g) + NO(g)

for temperatures below about 773 K involves the following steps -

2NO2 → NO3 + NO (slow) .....(1)

NO3 + CO → NO2 + CO2 (fast).....(2)

(a) Write the rate equation for the reaction if this mechanism is correct.

(b) Give the order of the reaction below 773 K with respect to

(i) NO2

(ii) CO

(c) At temperatures well above 773 K the probable mechanism for the complete reaction consists of only one step. Write the rate equation for the reaction if this mechanism is correct.

(d) Give the order of the reaction in (c) with respect to

(i) NO2

(ii) CO


At high temperatures, nitrogen oxide and hydrogen react according to the stoichiometric equation -

2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

A likely mechanism for the reaction involves the following series of steps -

NO(g) + NO(g) N2O2(g) fast, reversible (1)

N2O2(g) + H2(g) → N2O(g) + H2O(g) slow (2)

N2O(g) + H2(g) → N2(g) +H2O(g) fast (3)

What is the rate equation for the reaction if the above mechanism is correct?


A certain reaction has the following rate equation -

rate = k [H2][Cu2+]2 / [H+]

(a) What is the overall order of the reaction?

(b) What would be the effect on the rate of reaction if-

(i) The concentration of hydrogen were tripled, all other concentrations being held constant?

(ii) The concentrations of all three species in the rate equation were simultaneously doubled?


The rate of the reaction H2O2 + 2I- + 2H+ → I2 + 2H2O can be determined by using spectrophotometric means to measure the time taken, after mixing reactants, for [I2] to reach 1.00 x 10-5 M. It can be assumed (for suitable starting concentrations) that the rate is very nearly constant for this period.

(a) In an experiment in which [H2O2] = 0.010 M, [I-] = 0.010 M, and [H+] = 0.100 M, the time taken for [I2] to reach 1.00 x 10-5 M was 5.7 second. Calculate the rate of reaction (d[I2]/dt).

(b) When the experiment was repeated with [H2O2] = 0.0050 M, [I-] = 0.0100 M and [H+] = 0.100 M, the time was 11.5 second. Calculate the rate now.

(c) From these calculations show that the reaction rate depends on [H2O2] raised to the first power, i.e. that R ∝ [H2O2]

(d) Given the further information that the rate law is R = k [H2O2] [H+][I-], calculate the rate constant, k.

(e) Predict the rate of reaction if

[H2O2] = 0.0500 M, [H+] = 0.100 M and [I-] = 0.0200 M.


The thermal decomposition of nitrogen dioxide can be represented by the equation

2NO2(g) → 2NO(g) + O2(g)

From an experimental study, the following results were obtained for the rate of the reaction at 700 K.

Initial concentration of NO2

(mole litre-1)

Initial rate of reaction

(mole litre-1 second-1)





(a) Use this data to deduce a general expression for the rate of the reaction.

(b) What is the numerical value of the rate constant at 700 K?


For the reaction RCl + OH- → ROH + Cl- the following rate equations were determined experimentally for dilute solutions.


R = (CH3)3C-

R = CH3-

Rate = k1.[RCl]

Rate = k2.[RCl] x [OH-]

(k1 and k2 are rate constants)

For each reaction indicate by writing "yes" or "no" in each of columns whether the following possible reaction mechanisms are compatible with the rate equation.

Possible mechanism

R is (CH3)3C-

R is CH3-

RCl →R+ + Cl- (slow)

R+ + OH- → ROH (fast)


RCl →R+ + Cl- (fast)

R+ + OH- → ROH (slow)


RCl + OH- → ROH + Cl-

(one step)

Answers to Kinetics (Advanced) Questions  


(a) The rate law can only be determined by experiments such as those given in the data, and not from the coefficients in the stoichiometric equation.

In the experiment where the concentration of A (written as [A]) is not altered but the [B] is doubled, the result is no change in the rate of reaction. Hence the rate is not determined by [B]. i.e. the rate is zero order with respect to B.

In the experiment where both [A] and [B] are doubled, the rate increases by a factor of 4. However, as the rate is not affected by changes in [B], this increase in rate must be attributable entirely to the changed [A]. Thus doubling [A] causes a fourfold increase in rate, so the rate ∝ [A]2 or the reaction is second order with respect to A.

Thus the rate law is rate ∝ [A]2[B]0 or rate = k[A]2.

(b) The reaction being second order with respect to A means that in the rate determining step, two molecules of reactant A must combine to form an intermediate which could be represented as A.A and as this is the rate-determining step, it must be the slow step. In a subsequent step, one of these A.A intermediate species reacts with one molecule of B (from the stoichiometric equation) to form one molecule of the product, C. This second step must be fast as the rate is not affected by [B]. This mechanism could be shown by the following sequence of steps:

A + A → A.A (slow step)

A.A + B → C (fast step)


(a) In the slow (rate-determining) step (1), two molecules of NO2 combine to form the intermediate species, NO3 plus a molecule of one product, NO. The reaction is therefore second order with respect to NO2 or rate = k[NO2]2.

(b)(i) second order see (a) above

(ii) As CO does not feature as a reactant in the slow step but only appears in the fast step, the order of the reaction with respect to CO is zero.

(c) If the mechanism at the higher temperature is just a single step, then it must be the same as shown in the stoichiometric equation and so

rate ∝ [NO2][CO] or rate = k[NO2][CO].

(d) From the rate equation in part (c), the reaction is now first order with respect to both NO2 and CO.

From the slow step, rate ∝ [N2O2][H2].

The rate equation in terms of these species would be rate = k2[N2O2][H2]. However, one of these species, N2O2, is an intermediate produced by the equilibrium step 1 and not one of the original reactants whose concentrations can be measured before the reaction begins.

The [N2O2] can be obtained in terms of the concentrations of the initial reactants using the equilibrium constant for step 1 as follows:

Kc = [N2O2] / [NO]2

∴ [N2O2] = Kc x [NO]2

Substituting this in the rate equation rate = k2 [N2O2][H2] gives

Rate = k2Kc[NO]2[H2] or, combining the constants,

Rate = k[NO]2[H2].

This mechanism is an illustration of how a reaction can be third order overall while the rate determining step is only second order. In reactions, it is highly unlikely that more than two species will collide simultaneously with sufficient energy and with the required impact orientation for them to react, so a third order mechanistic step is highly improbable. So while this reaction mechanism would be third order overall, it is only second order in the slow step, (2).


(a) The overall order for the reaction is obtained by adding all the exponents of the concentrations in the rate equation.

The rate equation here is rate = k[H2][Cu2+]2[H+]-1,

so overall order = 1 + 2 +(-1) = 2, i.e. second order.

This is an example of a reaction where increasing the concentration of a reactant (the H+) causes the reaction to proceed more slowly as seen by its negative exponent in the rate equation. Thus the greater the concentration of H+, the more the reaction is inhibited.

(b)(i) Increasing the [H2] by a factor of three would increase the rate by a factor of 3 as the reaction is first order with respect to H2

(ii) Doubling [H2] doubles the rate as the reaction is first order with respect to H2. Doubling [Cu2+] increases the rate by a further factor of four as the reaction is second order with respect to Cu2+. Doubling [H+] will halve the rate as the order is -1 with respect to H+. Combining these changes, the overall change in rate is by a factor of 2 x 4 x = four times the original rate.


(a) The change in [I2] during the time interval of 5.7 second

= final concentration - initial concentration

= 1.00 x 10-5 - 0 M = 1.00 x 10-5 M.

∴ rate of formation of I2 = 1.00 x 10-5 / 5.7 M s-1 = 1.8 x 10-6 M s-1.

(b) The time interval is now 11.5 seconds to produce the same [I2].

∴ Rate = 1.00 x 10-5 / 11.5 = 8.7 x 10-7 M s-1.

(c) In the two experiments, only [H2O2] was altered. In the second experiment, halving its concentration as used in the first experiment caused the rate to halve, so the reaction must be directly proportional to [H2O2], or

Rate ∝ [H2O2]1.

(d) Given that the rate law is rate = k[H2O2][H+][I-], k can be calculated by substituting either of the sets of data given in (a) or (b) above.

Using data from (b), 8.7 x 10-7 = k (0.0050) (0.100) (0.010) = k x 5.0 x 10-6

k = 8.7 x 10-7 / 5.0 x 10-6 = 1.7 x 10-1 M-2 s-1.

(e) Using this value of k in the rate equation,

Rate = 1.8 x 10-1 (0.0500) (0.100) (0.0200)

= 1.8 x 10-5 M s-1.

Alternatively, one could use the method of direct comparison of this new data with that in (b) above and note that only [I-] is changed, being increased from 0.0100 M to 0.0200 M. Given the reaction is first order with respect to I-, this doubling of [I-] would double the rate calculated in (b) so now

rate = 2 x 8.7 x 10-7 = 1.7 x 10-5 M s-1.

6 (a) In this type of problem, check to see if the initial concentration data for the series of experiments has been changed by simple factors such as doubling. In this situation, the rate equation can be deduced merely by inspection.

However, the data given in this problem does not have a simple relationship between the initial concentrations used in the two experiments, so the deduction of the rate law is a little more complicated. Using the rate law, rate = k[NO2]n, substitution of the two sets of data results in two equations with two unknowns, k and n, which can therefore be obtained.

From the first experiment, 0.0031 = k (0.0225)n ................(1)

From the second experiment, 0.0016 = k (0.0162)n ..........(2)

Divide equation (1) by equation (2)

0.0031 / 0.0016 = k (0.0225)n /k (0.0162)n

1.94 = (1.39)n

Take logs of both sides: log (1.94) = n log(1.39)

0.286 = n x 0.143

∴ n = 2.0

and so the rate law is rate = k[NO2]2.

To evaluate k, substitute either set of data in the rate law. Using data from the first experiment,

0.0031= k (0.0225)2

k = 6.1 M-1 s-1.


When R is (CH3)3C-, the rate is directly proportional to [RCl], so only RCl should appear in the slow step of the mechanism equations for this reaction. When R is CH3-, the rate is proportional to both [RCl] and [OH-], so both RCl and OH- must appear in the slow step of any proposed mechanism for this reaction. Three proposed mechanisms are given.

RCl → R+ + Cl- (slow)

R+ + OH- → ROH (fast)

This mechanism is consistent with R being (CH3)3C- as only RCl appears in the slow step but not consistent with R being CH3-.

(ii) RCl → R+ + Cl- (fast)

R+ + OH- → ROH (slow)

In this proposed mechanism, the slow (rate-determining) step includes R+ which is not one of the initial reactants, so its concentration must be expressed in terms of [RCl] using the equilibrium constant for the first (fast) step as follows:

Kc = [R+][Cl-]/[RCl], so [R+] = Kc[RCl]/[Cl-]

The rate equation from the second step is rate = k[R+][OH-] which becomes

Rate = kKc[RCl][OH-]/[Cl-].

This rate equation requires the rate to be directly proportional to both [RCl] and [OH-] and inversely proportional to [Cl-]. (i.e. the presence of Cl- would inhibit the reaction.). This is not consistent with the observed rate data for either reaction and therefore is not a possible mechanism in either case.

(iii) RCl + OH- → ROH + Cl- (one step)

If the reaction mechanism consists of just one step, then this must be the rate-determining step. Here the rate equation would be

Rate = k[RCl][OH-] which is not consistent with the data when R is (CH3)3C- but would be consistent with the data for R = CH3-.